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# Principles of Chemistry I CHEM 1307

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This 37 page Class Notes was uploaded by Kara Dibbert on Thursday October 22, 2015. The Class Notes belongs to CHEM 1307 at Texas Tech University taught by Tamara Hanna in Fall. Since its upload, it has received 65 views. For similar materials see /class/226511/chem-1307-texas-tech-university in Chemistry at Texas Tech University.

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Date Created: 10/22/15

Gases and their Properties Pressure Experiments with gases involve measuring or controlling amount moles volume pressure and temperature Pressure P force area SI unit of pressure is the pascal Pa 1 newtonmeter2 1atm 760 mm Hg 760 torr 1013 kPa 1013 bar 1469 psi At sea level the barometric pressure is 760 mm Hg On top of the world s highest mountain the pressure is 270 mm Hg Boyle s Law The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure exerted by the gas V V As V decreases P increases V2 P1V1 or P1V1 PZV2 1000 900 800 70039 Pressure torr 600 500 l I J l l l l 7 8 9 1O 11 12 13 Volume mL Boyle s Law A sample of gaseous nitrogen in a 650 L automobile air bag has a pressure of 745 mm Hg If the sample is transferred to a 250 L bag at the same temperature what is the pressure in the bag Initial Conditions Final Conditions P1745 mm Hg P2mm Hg V1 650 L V2 250 L P1V1 PZV2 P2 P1V1 745 mm Hg X 504 1940 mm Hg V2 250 L Temperature and Volume To understand the relationship between V and T a student sealed a known amount of Hg in a capillary and measured the height of the Hg in the tube at different temperatures The amount of gas trapped in the tube remained constant temperature Height mm Ci J keivins 0 100 200 300 Temperature Charles s Law The experiment reveals V oc T As T increases V increases In equation form V CCT Charles s Law The volume of a fixed quantity of gas at constant pressure is directly proportional to absolute temperature CC y T When two measurements are taken on the same gas with a fixed amount at constant pressure V1 32 or V1T2 V2T1 T1 T2 Charles s Law Practice Problem A sample of CO2 in a gastight syringe has a volume of 250 mL at 200 C What is the final volume of the gas if you hold the syringe in your hand to raise its temperature to 37 C Convert all temperatures to kelvin V1T2 V2T1 v2 vsz 250 mL x 3104 265 mL T1 293 Think through the problem Temperature increased so volume must increase If V increases must multiply by a conversion factor larger than 1 The General Gas Law A combination of Charles s Law and Boyle s Law Use for problems involving changes in P T and V with a constant amount of gas VOCI VZCGI CGz P P T T1 T2 The volume of a fixed quantity of gas depends on the T and P It is not possible to state the V of gas without stating the T and P Standard Temperature and Pressure STP 0 C 273 K and 1 atm 760 torr General Gas Law Problem Suppose a helium balloon is launched when the temperature is 225 C and the pressure is 754 mm Hg If the volume of the balloon is 419 x 103 L will the volume be at a height of 20 miles when the pressure is 760 mm Hg and the temperature is 33 C P1V1 P2V2 T1 T2 v2 V1 x xL 419 x 103 L x 754 mm Fl x 240K p2 T1 760 mmd l 296K 337 x 104 L Does this make sense PlIVl The General Gas Law Charles s Law and Boyle s Law are special cases of the General Gas Law At Constant Pressure At Constant Temperature P1 P2 T1 T2 V1 la P1V1 PZV2 T1 T2 At Constant Volume v1v2 P1P2 T1 T2 A temperature increases pressure increases Avogadro s Hypothesis Equal volumes of all gases at the same temperature and pressure contain the same number of molecules At constant temperature and pressure V on n where n number of moles l I 1LHZ 1Lc2 ZLHCI H2 lg CI 9 a 2 HCIg u on awe H2 Clz 4 2 HCI What happens Avogadro s Hypothesis If you begin with 150 L of hydrogen gas what volume of nitrogen gas is required to complete the following reaction What is the theoretical yield of ammonia Assume all gases are at the same temperature and pressure N23H2 gt 2NH3 15mm x 1LN2 31 50 L N2 required 100 L NH3 formed 15mm x 2 2 3 The Ideal Gas Law Volume is proportional to all the measurable properties of gases The ideal gas law ties all the properties together with a single equation PV 2 nRT where R is the universal gas constant IDEAL GAS LAW PV nRT Fixed n my Fixed n and owd Pand T Boyle39s Law Charles s Law Avogadro s Law Voc 3 V 0 T v 00 n a 2mm mmsun Hmer Education Ideal Gas Law Constant At standard temperature and pressure 1 mol of gas occupies 22414 L of volume From these experimental data calculate the value of R PV nRT Rearrange to solve for R R nT RPXV latmx 22414L 00821L39atm nXT 1mo X 273K molK Ideal Gas Law Single Variable The hot air balloon used for the historic flight around France in 1783 was filled with 1300 moles of hydrogen If the temperature of the gas was 23 C and the pressure was 750 mm Hg what was the volume of the balloon PV nRT v nRT 1300 W 00821 L Mondrxx 296K P 0987 M 32 X 104 L Ideal Gas Law Molar Mass A useful variation of the ideal gas law is to replace the number of moles with the mass of a sample n mass molar mass PV mRT MM A 0105 9 sample of a gas has a pressure of 561 mmHg in a volume of 125 mL at 230 C What is its molar mass MM mRT 0105 g x 00821a12m moKx 296K 277 gmol PV 0738 yam 0125 Gas Density The density of a liquid or solid is expressed in gmL but for a gas the density is so low that the units are gL E d MMP V RT Calculate the density of CO2 at STP d MMP 44 gpao rx 1003 196 gL RT 00821 L tf pldl39KX 273K Because CO2 has a larger MM than air it has a higher density How many grams of CO2 would a 456 L flask contain at the temperature and pressure at which its density is 204 gL m VD 4561 x 204 g 930 g 1 Gas Stoichiometry at STP An air bag should be filled with gas at a pressure of 829 mm Hg at a temperature of 220 C The bag has a volume of 455 L What amount of sodium azide should be used to generate the required abount of gas 2 NaN3 s gt 2 Na 5 3 N29 Convert to moles of nitrogen gas n PV 109 emfx 455K 205 mol N2 RT 00821 J tr mol xx 295 x Use stoichiometry 205 x M x 6499 g NaN3 888 g NaN3 3 mom 1 moi Ham Gas Stoichiometw What volume of H2 measured at 739 torr and 21 C can be released by 427 9 Zn as it reacts with HCI Zn 2HCI gt ZnCl2 H2 First determine the number of moles of hydrogen 427gzrx M x 1 moi H 0653 moi H2 6538gZiT 1M Use the ideal gas law to solve v nRT 0653m6i x 00821 L vh Lm rK x 29 P 0972 M 162 L H2 Gas Mixtures and Partial Pressures When multiple gases are mixed ie air each gas exerts its own pressure called its partial pressure Dalton s Law of Partial Pressures The pressure of a mixture is the sum of the partial pressures of the different gases in the mixture Ptotal P1 P2 P3 Each gas behaves independently in the mixture in 279 mm Hg Ti Gas Mixtures and Partial Pressures The total pressure of a mixture of 150 g of halothane vapor CZHBrCIF3 and 235 g of 02 gas is 855 mm Hg What is the partial pressure of each gas PAV nART PBV nBRT Without knowing T and V how else can we solve the problem Mole Fraction X The mole fraction is the number of moles of a particular substance in a mixture divided by the total number of moles of all the substances present XA nA Totalquot Combine mole fraction with partial pressure PA XAPtotal Gas Mixtures and Partial Pressures The total pressure of a mixture of 150 g of halothane vapor CZHBrCIF3 and 235 g of 02 gas is 855 mm Hg What is the partial pressure of each gas 150 X 1 mol CgHBrCIF 00760 mol CZHBrCIF3 1974gC2HB1 39F 235963 X 1 mol 0 0734 mol 02 32013 8 Total number of moles 00760 mol 0734 mol 0810 mol Xhalothane nhalothane 0390760 mOI 0390938 ntotal 0810 mol Xowgen 13900 Xhalothane 0906 phalotheme 00938855 mm Hg 802 mm Hg p 0906855 mm Hg 7748 mm Hg oxygen Gas Mixtures and Partial Pressures Sodium metal reacts with water to produce hydrogen gas If the hydrogen gas is collected over water at a temperature of 25 C and a pressure of 756 mm Hg and occupies a volume of 774 L what is the number of grams of H2 formed The vapor pressure of water is 238 mm Hg at 25 C 2 Na 5 2 H20 I gt 2 NaOH aq H2 9 PHZ Ptotal PHZO 756 mm Hg 238 mm Hg 732 mm Hg How many moles of hydrogen are formed n PV 0963atr x 774Z 0305 mol H2 RT 00821 J tr mol x x 298 x 0305 X 2016 H 06159H2 Imam Kinetic Theow of Gases This theory describes the behavior of particles on the molecular level in a way we can visualize them on the macroscopic scale The distance between molecules in the gas phase is very large compared to the size of the particles themselves Particles of gas are in constant random and rapid motion The molecules collide with each other and the container walls but do not lose kinetic energy The average kinetic energy of gas molecules is proportional to the gas temperature All gases regardless of their molecular masses have the same average kinetic energy at the same temperature Pressure arises from collisions of the molecules with the container walls Molecular Speed and Kinetic Energy The molecules in a gas sample do not move at the same speed At a given temperature some molecules move fast some move slow and many move at an intermediate speed At 25 C more molecules are F moving at about 400 ms than at any other speed Many more molecules are moving at 02 at 25 c 1600 ms when the sample is at 1000 C than when itis at 25 C 02 at 1000 c Number of molecules gt 0 200 400 600 800 1000 1200 1400 1600 1800 Molecular speed ms Molecular Speed and Kinetic Energy As the temperature increases the most probable speed increases as does the number of molecules traveling at the highest speeds E Wm average We cannot calculate the KE for a collection of molecules because the speeds are different but we can calculate an average value Average KE is also directly proportional to temperature using the constant 32R where R 8314 Jmol K E 32RT We can relate molecular mass kgmol average speed and temp using Maxwell s equation also called the rootmeansquare speed Ju Z MM Molecular Speed and Kinetic Energy All gases have the same average KE at the same temperature However molecules of different gases do not have the same average speed 2 The smaller the mass 02 the greater the speed 2 The higher the temp the greater the speed HeLE MM 0 500 1000 1500 2000 Molecular speed ms Number of molecules gt Molecular Speed and Kinetic Energy Calculate the rootmeansquare speed of O2 molecules at 25 C xl MM J32 3x83145JrmKX298K 482 mS 320x10 3kgrml Remember to change the units of MM to kgmol 1 J 1 kngs2 Kinetic Molecular Theow and Gas Laws Pressure arises from collisions of gas molecules with the walls of the container holding the gas gas pressure force of collisions area The force depends on the number of collisions and the average force per collision As temperature increases the average KE increases the force of the collisions increases and more collisions occur per second because the speed of the gas molecules increases When this occurs the pressure increases P on T when n and V are constant Kinetic Molecular Theow and Gas Laws Increasing the number of molecules does not change the average collision force but does increase the number of collisions per second When this occurs the pressure increases P on n when T and V are constant When T or n increase and the pressure remains constant the volume must increase as does the area over which collisions can occur V or nT when P is constant When T and n remain constant while the volume of the container is decreased the number of collisions on the container per second must increase resulting in an increase in pressure P a 1V when T and n are constant Diffusion and Effusion Diffusion The gradual mixing of molecules of two or more gases due to their random molecular motions time gt Over time the gases will mix completely Gases with higher molecular masses diffuse more slowly Diffusion and Effusion The movement of gas through a tiny opening in a container into another container where the pressure is very low Before effusion During effusion 37 N 2 3 H2 0 0 19 Vacuum gal Q I Porous barrier Diffusion and Effusion Graham s Law The rate of effusion amount of gas that moves in a given amount of time is inversely related to the square root of the MM rate of effusion of gas A MMB time of effusion of gas B rate of effusion of gas B MM A time of effusion of gas A Tetrafluoroethylene C2F4 effuses through a barrier at a rate of 46 X 10396 molhr An unknown gas consisting of B and H effuses at a rate of 58 X 10396 molhr under the same conditions What is the molar mass of the unknown gas 46 X 10396 molhr MMunknown 58 X 10396 molhr 1000gmo Solving for the MM yields a value of 63 gmol Light molecules effuse more rapidly than heavy molecules Nonideal Behavior Real Gases Deviations from the ideal gas law occur at high pressures or low temperatures This is due to a breakdown of the assumptions used to describe an ideal gas Assumption Gas particles have negligible size At around 1 atm of pressure a single gas molecule is very small relative to the total volume of space it moves in Problem Higher pressures mean more gas particles When the pressure is increased to 1000 atm each gas particle has a much smaller share of the volume and the particle size becomes important Nonideal Behavior Real Gases Deviations from the ideal gas law occur at high pressures or low temperatures Assumption Gas particles behave independently Gas molecules do not interact with each other by any intermolecular forces Problem Gases can be liquefied therefore forces exist When a gas molecule is about to strike the container wall other molecules near it exert a slight pu so the molecule actually hits the wall with less force this causes the gas pressure for a real gas to be lower than expected van der Waals Equation The breakdown of the ideal gas equation can be corrected using the van der Waals equation P a Killzxv 93 nRT I l 39 correction for correCtlon for molecular volume Intermolecular forces observed pressure container volume The observed pressure is lower than the ideal pressure due to intermolecular forces so the correction factor is added The molecules take up more space so less volume is available to move around so the correction factor is subtracted Ideal Gas Vs Real Gas Consider a sample of 400 moles of CI2 gas in a 400 L tank at 100 C What is the pressure using the ideal gas law P nRT 400 maix 0082watmme1 Kx 373 V v 400 Pideal 306 atm What is the pressure using van der Waals equation a 649 b 00562 P a 0712v bn nRT Preal 260 atm

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